Find the [tex]\(x\)[/tex]-intercepts of the following function:
[tex]\[ y = -2x^2 + 16x - 30 \][/tex]

A. [tex]\((-6,0), (-10,0)\)[/tex]
B. [tex]\((3,0), (5,0)\)[/tex]
C. [tex]\((6,0), (10,0)\)[/tex]
D. [tex]\((-3,0), (-5,0)\)[/tex]



Answer :

To find the [tex]\(x\)[/tex]-intercepts of the quadratic function [tex]\( y = -2x^2 + 16x - 30 \)[/tex], we need to determine the values of [tex]\(x\)[/tex] when [tex]\(y = 0\)[/tex]. This requires solving the quadratic equation:

[tex]\[ -2x^2 + 16x - 30 = 0 \][/tex]

Let's go through the steps to find these intercepts:

### 1. Identify the Coefficients
First, identify the coefficients [tex]\(a\)[/tex], [tex]\(b\)[/tex], and [tex]\(c\)[/tex] in the quadratic equation [tex]\( ax^2 + bx + c = 0 \)[/tex]:
- [tex]\(a = -2\)[/tex]
- [tex]\(b = 16\)[/tex]
- [tex]\(c = -30\)[/tex]

### 2. Calculate the Discriminant
The discriminant [tex]\(\Delta\)[/tex] of a quadratic equation [tex]\(ax^2 + bx + c = 0\)[/tex] is given by:
[tex]\[ \Delta = b^2 - 4ac \][/tex]

Plug in the values of [tex]\(a\)[/tex], [tex]\(b\)[/tex], and [tex]\(c\)[/tex]:
[tex]\[ \Delta = 16^2 - 4(-2)(-30) \][/tex]
[tex]\[ \Delta = 256 - 240 \][/tex]
[tex]\[ \Delta = 16 \][/tex]

### 3. Use the Quadratic Formula
The [tex]\(x\)[/tex]-intercepts are given by the solutions to the quadratic equation, which can be found using the quadratic formula:
[tex]\[ x = \frac{-b \pm \sqrt{\Delta}}{2a} \][/tex]

Plug in [tex]\(a\)[/tex], [tex]\(b\)[/tex], and [tex]\(\Delta\)[/tex]:
[tex]\[ x = \frac{-16 \pm \sqrt{16}}{2(-2)} \][/tex]
[tex]\[ x = \frac{-16 \pm 4}{-4} \][/tex]

### 4. Calculate the Solutions
Now, solve for [tex]\(x\)[/tex] using the [tex]\(\pm\)[/tex] in the equation.

For the positive case:
[tex]\[ x = \frac{-16 + 4}{-4} \][/tex]
[tex]\[ x = \frac{-12}{-4} \][/tex]
[tex]\[ x = 3 \][/tex]

For the negative case:
[tex]\[ x = \frac{-16 - 4}{-4} \][/tex]
[tex]\[ x = \frac{-20}{-4} \][/tex]
[tex]\[ x = 5 \][/tex]

### 5. Conclusion
The [tex]\(x\)[/tex]-intercepts of the function [tex]\( y = -2x^2 + 16x - 30 \)[/tex] occur at:
[tex]\[ (x, y) = (3, 0) \quad \text{and} \quad (5, 0) \][/tex]

Therefore, the correct choice is:
[tex]\((3, 0)\)[/tex] and [tex]\((5, 0)\)[/tex]